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18 Dec 2023, 07:00
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12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun
A = {9, 10, 11, 12, x}
If the median and the average (arithmetic mean) of the data set A, shown above, are equal, what is the value of x ?
(1) x is not an integer
(2) x is not a prime number.
A = {9, 10, 11, 12, x}
If the median and the average (arithmetic mean) of the data set A, shown above, are equal, what is the value of x ?
(1) x is not an integer
(2) x is not a prime number.
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19 Dec 2023, 08:29
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GMAT Club Official Explanation:
A = {9, 10, 11, 12, x}
If the median and the average (arithmetic mean) of the data set A, shown above, are equal, what is the value of x ?
Observe that the given set, A, can have only three possible medians:
- 10, when \(x \leq 10\)
- x, when \(10 \leq x \leq 11\)
- 11, when \(11 \leq x\)
When the median is 10, equating the average and the median gives us (9 + 10 + 11 + 12 + x)/5 = 10, resulting in x = 8;
When the median is x, equating the average and the median gives (9 + 10 + 11 + 12 + x)/5 = x, resulting x = 10.5;
When the median is 11, equating the average and the median gives (9 + 10 + 11 + 12 + x)/5 = 11, resulting x = 13.
(1) x is not an integer
The only non-integer value among the three possible values of x is 10.5. Sufficient.
(2) x is not a prime number.
This excludes 13 and leaves two possible values for x: 8 and 10.5. Not sufficient.
Answer: A.
General Discussion
18 Dec 2023, 10:04
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\(average = \frac{42+x}{5}\)
We have two cases:
a) x is an integer
we take values:
- if x = 3 then average would be 9 but the median would be 10 so not possible
- if x = 8 then average would be 10 and the median would be 10 so possible
- if x = 13 then average would be 11 and the median would be 11 so possible
- if x = 18 then average would be 12 but the median would be 11 so not possible
For values of x less than 3 the median would be 10 and the average would be less than 9 so not possible
For values of x greater than 18 the median would be 11 and the average would be greater than 12 so not possible
So we have two possible integer values for x: 8 and 13
b) x is not an integer
In this case x must be the median because if x is a non integer, average is a non integer and the only non integer in the set to be the median is x.
\(\frac{42+x}{5} = x\)
\(x = 10.5\)
So we have one possible non integer value for x: 10.5
Total possible values: 8, 10.5, 13
1) Only 10.5 is not an integer
SUFFICIENT
2) 8 and 10.5 are not prime numbers
INSUFFICIENT
IMO A
We have two cases:
a) x is an integer
we take values:
- if x = 3 then average would be 9 but the median would be 10 so not possible
- if x = 8 then average would be 10 and the median would be 10 so possible
- if x = 13 then average would be 11 and the median would be 11 so possible
- if x = 18 then average would be 12 but the median would be 11 so not possible
For values of x less than 3 the median would be 10 and the average would be less than 9 so not possible
For values of x greater than 18 the median would be 11 and the average would be greater than 12 so not possible
So we have two possible integer values for x: 8 and 13
b) x is not an integer
In this case x must be the median because if x is a non integer, average is a non integer and the only non integer in the set to be the median is x.
\(\frac{42+x}{5} = x\)
\(x = 10.5\)
So we have one possible non integer value for x: 10.5
Total possible values: 8, 10.5, 13
1) Only 10.5 is not an integer
SUFFICIENT
2) 8 and 10.5 are not prime numbers
INSUFFICIENT
IMO A
